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If $4 \bar{i} + 7 \bar{j} + 8 \bar{k}, 2 \bar{i} + 3 \bar{j} + 4 \bar{k}$ and $2 \bar{i} + 5 \bar{j} + 7 \bar{k}$ are the position vectors of the vertices A, B and C of triangle ABC, the position vector of the point where the bisector of $∠ A$ meets BC is

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$\frac{2}{3} (- 6 \bar{i} - 8 \bar{j} - 6 \bar{k})$

$\frac{1}{3} (6 \bar{i} + 13 \bar{j} + 18 \bar{k})$

$2 (\bar{i} + \bar{j} + \bar{k})$

$\frac{2}{3} (6 \bar{i} + 8 \bar{j} + 6 \bar{k})$

answer is C.

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Detailed Solution

$\bar{B D} : \bar{D C} = A B : A C = 6 : 3 = 2 : 1 \Rightarrow \bar{O D} = \frac{(4 \bar{i} + 10 \bar{j} + 14 \bar{k}) + (2 \bar{i} + 3 \bar{j} + 4 \bar{k})}{3} = \frac{6 \bar{i} + 13 \bar{j} + 18 \bar{k}}{3}$

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